Answer
a.
$\left\{\begin{array}{l}
w+2x+5y+5z=3\\
w+x+3y+4z=-1\\
w-x-y+2z=3
\end{array}\right.$
b.
$\{(1-y-3z, -2-2y-z, y, z),\ \ y,z\in \mathbb{R}\}$
Work Step by Step
a.
The augmented matrix has 5 columns, so there are 4 variables.
Let the variables be w,x,y, and z.
System of equations:
$\left\{\begin{array}{l}
w+2x+5y+5z=3\\
w+x+3y+4z=-1\\
w-x-y+2z=3
\end{array}\right.$
b.
The reduced system: $\left\{\begin{array}{l}
w+y+3z=1\\
x+2y+z=-2
\end{array}\right.$
Taking y and z as parameters,
express w and x in terms of y and z.
$w+y+3z=1 \Rightarrow w=1-y-3z$
$x+2y+z=-2 \Rightarrow x=-2-2y-z$
The solution set is
$\{(1-y-3z, -2-2y-z, y, z),\ \ y,z\in \mathbb{R}\}$.
